Question

pH of a solution by mixing 50 mL of 0.1 M $Na_3PO_4$ and 50 mL of 0.05 M $NaH_2PO_4$ is : [For $H_3PO_4$: $pK_1$ = 2.1, $pK_2$ = 7.2, $pK_3$ = 12 and log 3 = 0.3]

• A. 9.6
• B. 4.65
• C. 7.2
• D. 11.7

Answer 1

Answer: (D)

11.7

Answer 2

The initial solution is formed by mixing 50 mL of 0.1 M $Na_3PO_4$ and 50 mL of 0.05 M $NaH_2PO_4$. The total volume of the mixed solution is 50 mL + 50 mL = 100 mL = 0.1 L.

Calculate the initial moles of the phosphate species:

Moles of $Na_3PO_4 = \text{Volume} \times \text{Concentration} = 0.05 \text{ L} \times 0.1 \text{ mol/L} = 0.005 \text{ mol}$. $Na_3PO_4$ dissociates completely into $3Na^+$ and $PO_4^{3-}$. So, moles of $PO_4^{3-} = 0.005 \text{ mol}$.

Moles of $NaH_2PO_4 = \text{Volume} \times \text{Concentration} = 0.05 \text{ L} \times 0.05 \text{ mol/L} = 0.0025 \text{ mol}$. $NaH_2PO_4$ dissociates completely into $Na^+$ and $H_2PO_4^-$. So, moles of $H_2PO_4^- = 0.0025 \text{ mol}$.

The solution contains $H_2PO_4^-$ ions and $PO_4^{3-}$ ions. These species can react via proton transfer:

$H_2PO_4^- + PO_4^{3-} \rightleftharpoons HPO_4^{2-} + HPO_4^{2-}$

Let's consider the relative acidity and basicity of the initial species.

$H_2PO_4^-$ can act as an acid ($K_2 = 10^{-7.2}$) or a base ($K_b = K_w/K_1 = 10^{-11.9}$). It is primarily acidic.

$PO_4^{3-}$ is a strong base (conjugate of $HPO_4^{2-}$). $PO_4^{3-} + H_2O \rightleftharpoons HPO_4^{2-} + OH^-$, $K_b = K_w/K_3 = 10^{-14}/10^{-12} = 10^{-2}$.

The reaction $H_2PO_4^- + PO_4^{3-} \rightleftharpoons 2 HPO_4^{2-}$ is a reaction between an acid ($H_2PO_4^-$) and a base ($PO_4^{3-}$). The equilibrium constant for this reaction is $K = \frac{[HPO_4^{2-}]^2}{[H_2PO_4^-][PO_4^{3-}]}$. This reaction can be seen as the sum of $H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}$ ($K_2$) and $PO_4^{3-} + H^+ \rightleftharpoons HPO_4^{2-}$ ($1/K_3$). So, $K = K_2 \times (1/K_3) = 10^{-7.2} \times (1/10^{-12}) = 10^{-7.2 + 12} = 10^{4.8}$.

Since the equilibrium constant is large ($10^{4.8}$), the reaction proceeds significantly towards the formation of $HPO_4^{2-}$. We can assume the reaction goes nearly to completion, limited by the reactant present in the smaller amount.

Initial moles: $H_2PO_4^- = 0.0025$ mol, $PO_4^{3-} = 0.005$ mol.

$H_2PO_4^-$ is the limiting reactant.

Assume complete reaction of $H_2PO_4^-$:

Moles of $H_2PO_4^-$ reacted = 0.0025 mol.

Moles of $PO_4^{3-}$ reacted = 0.0025 mol.

Moles of $HPO_4^{2-}$ formed = $2 \times 0.0025$ mol = 0.005 mol.

After the reaction goes to near completion:

Moles of $H_2PO_4^-$ remaining = $0.0025 - 0.0025 = 0$ mol.

Moles of $PO_4^{3-}$ remaining = $0.005 - 0.0025 = 0.0025$ mol.

Moles of $HPO_4^{2-}$ formed = $0.005$ mol.

The resulting solution contains $HPO_4^{2-}$ and $PO_4^{3-}$. This is a buffer solution, as $PO_4^{3-}$ is the conjugate base of the weak acid $HPO_4^{2-}$.

The relevant equilibrium is the third dissociation step of $H_3PO_4$:

$HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}$

The acid is $HPO_4^{2-}$ and its conjugate base is $PO_4^{3-}$. The $pK_a$ for this equilibrium is $pK_3 = 12$.

We can use the Henderson-Hasselbalch equation for this buffer system:

pH = $pK_a + \log \frac{[conjugate \, base]}{[acid]}$

pH = $pK_3 + \log \frac{[PO_4^{3-}]}{[HPO_4^{2-}]}$

The concentrations of the species in the 100 mL (0.1 L) solution are:

$[PO_4^{3-}] = \frac{0.0025 \text{ mol}}{0.1 \text{ L}} = 0.025 \text{ M}$.

$[HPO_4^{2-}] = \frac{0.005 \text{ mol}}{0.1 \text{ L}} = 0.05 \text{ M}$.

Substitute these values into the Henderson-Hasselbalch equation:

pH = $12 + \log \frac{0.025}{0.05}$

pH = $12 + \log (0.5)$

pH = $12 + \log (1/2)$

pH = $12 + \log 1 - \log 2$

pH = $12 + 0 - \log 2$

pH = $12 - \log 2$.

The question provides log 3 = 0.3. This value is not directly used in the calculation $12 - \log 2$. There might be a typo in the provided information or the options. However, assuming the calculation $12 - \log 2$ is correct based on the species present and relevant $pK_a$, we need the value of log 2. A common approximation is log 2 ≈ 0.301. If we use log 2 ≈ 0.3 (as suggested by the value of log 3 provided), then:

pH = $12 - 0.3 = 11.7$.

This value matches option (D). Let's verify if there's any scenario where log 3 is used. The ratio of concentrations is 0.025/0.05 = 1/2. No 3 involved here. The pK values are 2.1, 7.2, 12. No direct use of log 3 is apparent in the buffer calculation involving pK3.

Let's re-check the calculation steps. The identification of the buffer system ($HPO_4^{2-}/PO_4^{3-}$) and the use of $pK_3$ seem correct based on the initial amounts and the strong reaction between the initial species. The Henderson-Hasselbalch equation is applied correctly. The final calculation depends on the value of log 2. Assuming the question implies log 2 ≈ 0.3, the result is 11.7.

Let's consider the possibility of a typo in the question, e.g., if the concentration ratio was 1:3 or 3:1, then log 3 would be used. Or if the relevant pKa was related to log 3.

Given the options and the provided log 3 = 0.3, the calculation pH = 12 - log 2 = 12 - 0.3 = 11.7 seems the most plausible path to one of the options, assuming log 2 ≈ log 3 ≈ 0.3 in this context.

Final check of the buffer system. The average protonation state of the phosphate species is $(0.005 \times 0 + 0.0025 \times 2) / (0.005 + 0.0025) = 0.005 / 0.0075 = 2/3$. An average protonation state of 2/3 lies between 0 (for $PO_4^{3-}$) and 1 (for $HPO_4^{2-}$). This confirms that the major species at equilibrium should be $PO_4^{3-}$ and $HPO_4^{2-}$, forming a buffer system governed by $pK_3$. The ratio of moles of $HPO_4^{2-}$ to $PO_4^{3-}$ should give the average protonation state of 2/3. Let the moles be $n_2$ ($HPO_4^{2-}$) and $n_3$ ($PO_4^{3-}$). Total moles $n_2 + n_3 = 0.0075$. Total protons $n_2 \times 1 + n_3 \times 0 = n_2$. Average protons per phosphate = $n_2 / (n_2 + n_3) = 2/3$. So $n_2 = (2/3)(n_2 + n_3)$. This is not directly solving for $n_2$ and $n_3$ from the average.

Let's use the total proton count approach from the similar question.

Total moles of phosphate = 0.005 + 0.0025 = 0.0075 mol.

Initial proton count on phosphate species relative to $PO_4^{3-}$ (0 protons): $0.005 \times 0 + 0.0025 \times 2 = 0.005$ proton units.

At equilibrium, the species are $HPO_4^{2-}$ (1 proton) and $PO_4^{3-}$ (0 protons). Let the moles be $n_2$ and $n_3$.

Total phosphate: $n_2 + n_3 = 0.0075$.

Total proton units: $n_2 \times 1 + n_3 \times 0 = n_2$.

Equating initial and equilibrium proton units: $n_2 = 0.005$.

Then $n_3 = 0.0075 - n_2 = 0.0075 - 0.005 = 0.0025$.

This confirms the moles of $HPO_4^{2-}$ and $PO_4^{3-}$ at equilibrium are 0.005 mol and 0.0025 mol respectively. The concentrations are 0.05 M and 0.025 M.

The buffer calculation pH = $12 + \log (0.025/0.05) = 12 - \log 2$ is correct.

Assuming log 2 = 0.3, pH = 11.7.